i have a little math/coding problem witch i don`t have any idea how could i do it work in a simple way, so the problem is is need to make a line shorter, with 15
in my program i have :
http://gyazo.com/aff5ff61fb9ad3ecedde3118d9c0895e
the line takes the center coordinates of both circles and draws from one to another, but i need it to be from the circumference of the circles, so it wont get inside
the code im using is :
var line:Shape = new Shape();
line.graphics.lineStyle(3,0xFF0000,2);
line.graphics.moveTo(sx,sy);
line.graphics.lineTo(fx,fy);
this.addChild(line);
arrow2(sx,sy,fx,fy);
var rline:Shape = new Shape();
rline.graphics.lineStyle(3,0xFF0000,2);
rline.graphics.moveTo(fx,fy);
rline.graphics.lineTo(xa,ya);
this.addChild(rline);
var rline2:Shape = new Shape();
rline2.graphics.lineStyle(3,0xFF0000,2);
rline2.graphics.moveTo(fx,fy);
rline2.graphics.lineTo(xb,yb);
this.addChild(rline2);
the rline and rline2 function is for the arrow lines, now my question is how do i make it shorter not depending on it direction so it will not overlap the circle
You can use vectors to solve your problem; they're pretty easy to get the hang of, and pretty much indispensable for things like game dev or what you're trying to do. You can get an overview here: http://www.mathsisfun.com/algebra/vectors.html or by searching "vector math" in google
So first step is to get a vector from one circle to another (pretty much what you've done):
var vector:Point = new Point( circle2.x - circle1.x, circle2.y - circle1.y );
var length:Number = vector.length; // store the length of the vector for later
This is the equivalent of saying "if you start at circle1 and move along vector, you'll arrive at circle2"
Next thing we're going to do is normalise it; all this does is set the length to 1 - the direction is unchanged - this makes it easier to work with for what you're looking to do. A vector with length 1.0 is called a unit vector:
vector.normalize( 1.0 ); // you can pass any length you like, but for this example, we'll stick with 1.0
Now, to draw a line from one circle to another, but starting from the outside, we simply find the start and the end points. The starting point is simple the position of circle1 plus vector (normalised to unit length) multiplied by the radius of circle1:
var sx:Number = circle1.x + vector.x * circle1.radius; // or circle1.width * 0.5 if you don't store the radius
var sy:Number = circle1.y + vector.y * circle1.radius;
The ending point can be found by starting at our start point, and continuing along our vector for a distance equal to the distance between the two circles (minus their radii). The length value that we created earlier is the distance between your two circles, from one center point to another, so we can use that to get the distance minus the radii:
var dist:Number = length - ( circle1.radius + circle2.radius ); // or circle1.width * 0.5 etc
And so the end point:
var ex:Number = sx + vector.x * dist;
var ey:Number = sy + vector.y * dist;
And to draw the line between them:
var line:Shape = new Shape;
line.graphics.lineStyle( 1.0, 0x000000 );
line.graphics.moveTo( sx, sy );
line.graphics.lineTo( ex, ey );
this.addChild( line )
Related
i have polygon say (Hexagonal with 6 lines) this Hexagonal connected from center with 6 point That make 6 triangles
I need when move any point(cause to move triangles) ,, other points move like this point i mean if the left point move to lift other points move to the left and so on
the code I want like this ptcP1.x and ptcP1.y the point that i moving it other point move depend on ptcP1 movement note that, this equations work fine in square shape ,, put in Penta and hexa ..etc this equations in valid so can any one help me
function button1_triggeredHandler( event:Event ):void
{
mode="mode2";
//trace(list.selectedIndex);
if(list.selectedIndex==1)
{
DrawSqure.ptcP1.x = Math.random() + 50;
DrawSqure.ptcP1.y = Math.random() + 50;
DrawSqure.ptcP2.y = 50-DrawSqure.ptcP1.x;
DrawSqure.ptcP2.x = DrawSqure.ptcP1.y;
DrawSqure.ptcP3.x = 50-DrawSqure.ptcP1.y;
DrawSqure.ptcP3.y = DrawSqure.ptcP1.x;
DrawSqure.ptcP4.x = 50-DrawSqure.ptcP1.x;
DrawSqure.ptcP4.y = 50-DrawSqure.ptcP1.y;
}
As stated in the comments, storing the vertices/points into a container (Array or Vector) and then adjusting those positions when you move is the best way to do it. Here is an example of how that might work:
//setup array or vector of vertices
var polygonVertices:Array = [DrawPolygon.ptcP1, DrawPolygon.ptcP2, DrawPolygon.ptcP3, DrawPolygon.ptcP4];
This method will take all the vertices and apply the translation:
//function for adjusting all the vertices based on the distance you pass
function moveShape( vertices:Array, dx:Number, dy:Number ) {
var i:int;
for ( ; i < vertices.length; i++ ) {
vertices[i].x += dx;
vertices[i].y += dy;
}
}
Then you would just need to know your distance X & Y your shape has moved and you can call moveShape( polygonVertices, 100, 100 );
I inserted 100,100 as the distance parameters as an example, but this should give you the results you are looking for.
I have a Flash app where I am performing a scale and rotation operation about the center of _background:MovieClip (representing a page of a book). I have simple event listeners on the GESTURE_ROTATE and GESTURE_SCALE events of this MC which update some variables currentRotation and currentScaleX, currentScaleY. I then have the following code trigger on the ENTER_FRAME event of the app.
The problem I am encountering is upon rotating the MC beyond the limits of roughly 60 or -60 degrees, or scaling slightly and rotating, the MC begins to oscillate and finally spin wildly out of control and off the screen. I've tried several things to debug it, and even tried Math.flooring the currentRotationValue and rounding the values of currentScaleX/Y to the tenths place (Math.floor(currentScale * 10) / 10), but neither of these seems to remedy it. I'm a little stuck at this point and have tried researching as much as I can, but couldn't find anything. Any suggestions? Is there an issue with doing this operation on each frame perhaps?
private function renderPage(e:Event) {
var matrix:Matrix = new Matrix();
// Get dimension of current rectangle.
var rect:Rectangle = _background.getBounds(_background.parent);
// Calculate the center.
var centerX = rect.left + (rect.width/2);
var centerY = rect.top + (rect.height/2);
// Translating to the desired reference point.
matrix.translate(-centerX, -centerY);
matrix.rotate(currentRotation / 180) * Math.PI);
matrix.scale(currentScaleX, currentScaleY);
matrix.translate(centerX, centerY);
_background.transform.matrix = matrix;
}
I'm not certain what behaviour you're trying to produce, but I think the problem is that centerX and centerY define the middle of _background in _background.parent's coordinate space. You're then translating the matrix so that _background is rotated around the values centerX, centerY, but in _background's coordinate space.
Assuming you want _background to rotate around a point which remains static on screen, what you actually need to do is use two different Points:
matrix.translate(-_rotateAroundPoint.x, -_rotateAroundPoint.y);
matrix.rotate(currentRotation / 180) * Math.PI);
matrix.scale(currentScaleX, currentScaleY);
matrix.translate(_centerOnPoint.x, _centerOnPoint.y);
Where _rotateAroundPoint is the point around which _background should turn in it's own coordinate space, and _centerOnPoint is the point around which it should turn in its parent's coordinate space.
Both of those values only need to be recalculated when you want to pan _background, rather than every frame. For example:
private var _rotateAroundPoint:Point = new Point(_background.width * 0.5, _background.height * 0.5);
private var _centerOnPoint:Point = new Point(50, 50);
private function renderPage(e:Event) {
var matrix:Matrix = new Matrix();
matrix.translate(-_rotateAroundPoint.x, -_rotateAroundPoint.y);
matrix.rotate((currentRotation / 180) * Math.PI);
matrix.scale(currentScaleX, currentScaleY);
matrix.translate(_centerOnPoint.x, _centerOnPoint.y);
_background.transform.matrix = matrix;
}
I'm trying to zoom a DisplayObject into a certain point. I figured it would be easy, but I've spent a day now trying to figure it out.
Basically, I think this should work. Emphasis on should.
//newPoint is the point being centered. There is no initial scaling, so I do not need to compensate for that (yet)
//scale is the zoom level
//container is the parent of the obj
//obj is the object being scaled/panned
var p:Point = new Point(
( this.container.width - this.obj.width * scale + newPoint.x * scale ) / 2,
( this.container.height - this.obj.height * scale + newPoint.y * scale ) / 2
);
this.obj.scaleX = this.obj.scaleY = scale;
this.obj.x = p.x;
this.obj.y = p.y;
It centers the point if scale is 1, but it gets further and further away from center as you increase the scale. I've tried dozens of different methods. This method, which I have seen on several sites, produced the same exact results. Anyone have any idea how to get this to work?
EDIT 10-1-12:
As a followup, I took the code snippet that LondonDrugs_MediaServices provided as a basis for my original issue. I needed to be able to zoom to a specific point at a specific scale relative to the unscaled image (think how Google Maps zooms to a specific location). To do this, I had to center my image on the point before running the translation code. I've posted the additional code below. For other uses (pinch to zoom, scrolling, and double click), I used the code provided by Vesper, which worked quite well.
//obj is the object being translated
//container is its parent
//x and y are the coordinates to be zoomed to, in untranslated scaling
//obj.scaleX and obj.scaleY are always identical in my class, so there is no need to account for that
//calculates current center point, with scaling
var center:Point = new Point( ( this.container.width - this.obj.width * this.obj.scaleX ) / 2, ( this.container.height - this.obj.height * this.obj.scaleX ) / 2 );
//calulcates the distance from center the point is, with scaling
var distanceFromCenter:Point = new Point( this.obj.width * this.obj.scaleX / 2 - x * this.obj.scaleX, this.obj.height * this.obj.scaleX / 2 - y * this.obj.scaleX );
//center the object on that specific point
this.obj.x = center.x + distanceFromCenter.x;
this.obj.y = center.y + distanceFromCenter.y;
var mat:Matrix=new Matrix();
mat.translate(-p.x,-p.y);
mat.scale(desiredScale,desiredScale);
mat.translate(p.x,p.y);
yourObject.transform.matrix=mat;
The core point is that scaling is done around (0,0), but you can do it with matrix that describes affine transformations. You first make an empty matrix (that is, a matrix that doesn't transform), then apply a set of transformations to it. First, place a desired point at (0,0) by translating by -1*coordinates, then scale, then translate back.
hie guys....
thank's your comments...
i found the answer...
code :
gambar.addEventListener(TransformGestureEvent.GESTURE_ZOOM , onZoom);
function onZoom(event:TransformGestureEvent):void {
var locX:Number=event.localX;
var locY:Number=event.localY;
var stX:Number=event.stageX;
var stY:Number=event.stageY;
var prevScaleX:Number=gambar.scaleX;
var prevScaleY:Number=gambar.scaleY;
var mat:Matrix;
var externalPoint=new Point(stX,stY);
var internalPoint=new Point(locX,locY);
gambar.scaleX *= event.scaleX;
gambar.scaleY *= event.scaleY;
if(event.scaleX>1 && gambar.scaleX>6){
gambar.scaleX=prevScaleX;
gambar.scaleY=prevScaleY;
}
if(event.scaleY>1 && gambar.scaleY>6){
gambar.scaleX=prevScaleX;
gambar.scaleY=prevScaleY;
}
if(event.scaleX<1 && gambar.scaleX<0.8){
gambar.scaleX=prevScaleX;
gambar.scaleY=prevScaleY;
}
if(event.scaleY<1 && gambar.scaleY<0.8){
gambar.scaleX=prevScaleX;
gambar.scaleY=prevScaleY;
}
mat=gambar.transform.matrix.clone();
MatrixTransformer.matchInternalPointWithExternal(mat,internalPoint,externalPoint);
gambar.transform.matrix=mat;
}
The matrix answer is absolutely correct, but if you happen to be a Club GreenSock member you can get some nice functionality with very simple code with the TransformAroundPointPlugin
http://www.greensock.com/as/docs/tween/com/greensock/plugins/TransformAroundPointPlugin.html
You can see an example in the plugin explorer here:
http://www.greensock.com/tweenlite/#plugins
I use this to tween all my zooms and have much better performance than when I tried to do this manually. IMO the whole library is worth it's weight in gold (and no I have no connection other than liking the library). If you need any of the other features I'd look into it. It also has the ThrowProps plugin ( http://www.greensock.com/throwprops/ )which is very important if you are going to have a bounding box on mobile that you want to have a smooth return into the boundaries.
Set obj.x to -p.x and obj.y to -p.y, set the container scaleX and scaleY to the desired value and add p.x to the container x and p.y to the container y. Done!
I am trying to find out the Y position of a rotated object on stage, when only the X position is known. I am not extremely formiliar with how I'd go about doing this, but I know it'll be related to the Rotation of the border object.
What I need to do is know that based on the below X position that is worked out, what the exact maximum Y position can be before it hits the black border that is onscreen. I know that the original position is 280, but I am not sure at all how I then work out what the Y position is further down the line.
I have attached the code to find the X (all be it, it doesn't take into account any rotation as on this level it isn't needed), as well as a screenshot so you can understand clearly.
Thank you for your time.
private function init(e:Event = null):void{
var maxX:int = stage.width
var freeSpace:int = 300
var startX:int = Math.ceil(Math.random() * (maxX - (freeSpace+this.width))) + freeSpace;
this.x = startX
}
I'm not entirely sure on your question but hopefully these suggestions will help:
You can use the localToGlobal() function on a display object to return a rotated, translated, and scaled point within that display container to the stage. Example, $p:Point = myMovieClip.localToGlobal(new Point(10, 10));
A Matrix is also a nice and easy way to rotate a point. Example, var $mtx:Matrix = new Matrix(); $mtx.tx = 10; $mtx.ty = 10; $mtx.rotate(); and now $mtx.tx and $mtx.ty have the rotated result
Those probably won't answer your question, but I figured I'd mention them just in case and before I get into something more complex. Like wvxvw said you can't really solve the equation you're trying to do without some other variables. I wrote some code that shows how to find Y when comparing X to a point in a line segment:
import flash.display.Shape;
import flash.geom.Point;
import flash.display.Graphics;
import flash.events.MouseEvent;
var $s:Shape = new Shape();
addChild($s);
var borderStart:Point = new Point(stage.stageWidth/2, stage.stageHeight/2);
var borderRotation:Number = 45;
var borderLength:Number = 800;
var borderRad:Number = borderRotation * (Math.PI/180);
var borderEnd:Point = new Point(borderStart.x + Math.cos(borderRad) * borderLength, borderStart.y + Math.sin(borderRad) * borderLength);
stage.addEventListener(MouseEvent.MOUSE_MOVE, update);
function update(e:MouseEvent):void{
var $g:Graphics = $s.graphics;
$g.clear();
//Drawing the rotated border
$g.lineStyle(3, 0xff0000, .5);
$g.moveTo(borderStart.x, borderStart.y);
$g.lineTo(borderEnd.x, borderEnd.y);
//Finding if and where mouseX collides with our border
if (stage.mouseX >= Math.min(borderStart.x, borderEnd.x) && stage.mouseX <= Math.max(borderStart.x, borderEnd.x)){
var $x:Number = stage.mouseX;
//SOLVING HERE : Solve collision with X
var $percent:Number = ($x - borderStart.x)/(borderLength * Math.cos(borderRad));
var $y:Number = borderStart.y + Math.sin(borderRad) * borderLength * $percent;
//Drawing to our collision
$g.lineStyle(1, 0xffff00, .6);
$g.moveTo($x, 0);
$g.lineTo($x, $y);
$g.lineStyle(2, 0xffff00, 1);
$g.drawCircle($x, $y, 3);
trace("----\nCollision #\t" + "x: " + $x + "\ty:" + Math.round($y));
}
}
Hopefully this will give some insight on how to solve your particular issue.
I'm not sure if I'm answering the right question, because as you worded it, it's impossible to solve, or rather you would have to accept that Y can be just anything... (In order to be able to find a point in a vector space over R^2 you need a basis of two vectors of a form (x,y), but you only have a vector in R^1).
But it looks like you want to find an intersection of the "black line on the screen" - i.e. an arbitrary line and a vertical line through the lowest point of the "shape" which you want to fit. It's hard to tell from the question, what shape are you trying to fit, but if it is a rectangle, which is not rotated, then it would be either its bottom right or bottom left corner. You can then find which point to choose by comparing the angle between a horizontal line and the "black line" and the horizontal line and the bottom of the rectangle.
Next, you would need to find an intersection between these two lines, the formula can be found here: http://en.wikipedia.org/wiki/Line_intersection
If I create a rectangle with 100px width and 100px height and then rotate it, the size of the element's "box" will have increased.
With 45 rotation, the size becomes about 143x143 (from 100x100).
Doing sometimes like cos(angleRad) * currentWidth seems to work for 45 rotation, but for other bigger angles it doesn't.
At the moment I am doing this:
var currentRotation = object.rotation;
object.rotation = 0;
var normalizedWidth = object.width;
var normalizedHeight = object.height;
object.rotation = currentRotation;
Surely, there must be a better and more efficient way. How should I get the "normalized" width and height of a displayobject, aka the size when it has not been rotated?
The best approach would probably be to use the code posted in the question - i.e. to unrotate the object, check its width, and then re-rotate it. Here's why.
First, simplicity. It's obvious what's being done, and why it works. Anyone coming along later should have no trouble understanding it.
Second, accuracy. Out of curiosity I coded up all three suggestions currently in this thread, and I was not really surprised to find that for an arbitrarily scaled object, they give three slightly different answers. The reason for this, in a nutshell, is that Flash's rendering internals are heavily optimized, and among other things, width and height are not stored internally as floats. They're stored as "twips" (twentieths of a pixel) on the ground that further accuracy is visually irrelevant.
Anyway, if the three methods give different answers, which is the most accurate? For my money, the most correct answer is what Flash thinks the width of the object is when it's unrotated, which is what the simple method gives us. Also, this method is the only one that always give answers rounded to the nearest 1/20, which I surmise (though I'm guessing) to mean it's probably equal to the value being stored internally, as opposed to being a calculated value.
Finally, speed. I assume this will surprise you, but when I coded the three methods up, the simple approach was the fastest by a small margin. (Don't read too much into that - they were all very close, and if you tweak my code, a different method might edge into the lead. The point is they're very comparable.)
You probably expected the simple method to be slower on the grounds that changing an object's rotation would cause lots of other things to be recalculated, incurring overhead. But all that really happens immediately when you change the rotation is that the object's transform matrix gets some new values. Flash doesn't really do much with that matrix until it's next time to draw the object on the screen. As for what math occurs when you then read the object's width/height, it's difficult to say. But it's worth noting that whatever math takes place in the simple method is done by the Player's heavily optimized internals, rather than being done in AS3 like the algebraic method.
Anyway I invite you to try out the sample code, and I think you'll find that the simple straightforward method is, at the least, no slower than any other. That plus simplicity makes it the one I'd go with.
Here's the code I used:
// init
var clip:MovieClip = new MovieClip();
clip.graphics.lineStyle( 10 );
clip.graphics.moveTo( 12.345, 37.123 ); // arbitrary
clip.graphics.lineTo( 45.678, 29.456 ); // arbitrary
clip.scaleX = .87; // arbitrary
clip.scaleY = 1.12; // arbitrary
clip.rotation = 47.123; // arbitrary
// run the test
var iterations:int = 1000000;
test( method1, iterations );
test( method2, iterations );
test( method3, iterations );
function test( fcn:Function, iter:int ) {
var t0:uint = getTimer();
for (var i:int=0; i<iter; i++) {
fcn( clip, i==0 );
}
trace(["Elapsed time", getTimer()-t0]);
}
// the "simple" method
function method1( m:MovieClip, traceSize:Boolean ) {
var rot:Number = m.rotation;
m.rotation = 0;
var w:Number = m.width;
var h:Number = m.height;
m.rotation = rot;
if (traceSize) { trace([ "method 1", w, h ]); }
}
// the "algebraic" method
function method2( m:MovieClip, traceSize:Boolean ) {
var r:Number = m.rotation * Math.PI/180;
var c:Number = Math.abs( Math.cos( r ) );
var s:Number = Math.abs( Math.sin( r ) );
var denominator:Number = (c*c - s*s); // an optimization
var w:Number = (m.width * c - m.height * s) / denominator;
var h:Number = (m.height * c - m.width * s) / denominator;
if (traceSize) { trace([ "method 2", w, h ]); }
}
// the "getBounds" method
function method3( m:MovieClip, traceSize:Boolean ) {
var r:Rectangle = m.getBounds(m);
var w:Number = r.width*m.scaleX;
var h:Number = r.height*m.scaleY;
if (traceSize) { trace([ "method 3", w, h ]); }
}
And my output:
method 1,37.7,19.75
Elapsed time,1416
method 2,37.74191378925391,19.608455916982187
Elapsed time,1703
method 3,37.7145,19.768000000000004
Elapsed time,1589
Surprising, eh? But there's an important lesson here about Flash development. I hereby christen Fen's Law of Flash Laziness:
Whenever possible, avoid tricky math by getting the renderer to do it for you.
It not only gets you done quicker, in my experience it usually results in a performance win anyway. Happy optimizing!
Here's the algorithmic approach, and its derivation.
First, let's do the opposite problem: Given a rectangle of unrotated width w, unrotated height h, and rotation r, what is the rotated width and height?
wr = abs(sin(r)) * h + abs(cos(r)) * w
hr = abs(sin(r)) * w + abs(cos(r)) * h
Now, try the problem as given: Given a rectangle of rotated width wr, rotated height hr, and rotation r, what is the unrotated width and height?
We need to solve the above equations for h and w. Let c represent abs(cos(r)) and s represent abs(sin(r)). If my rusty algebra skills still work, then the above equations can be solved with:
w = (wr * c - hr * s) / (c2 - s2)
h = (hr * c - wr * s) / (c2 - s2)
You should get the bounds of your square in your object's coordinate space (which means no rotations).
e.g.
var b:Sprite = new Sprite();
b.graphics.lineStyle(0.1);
b.graphics.drawRect(0,0,100,100);
b.rotation = 10;
trace('global coordinate bounds: ' + b.getBounds(this));//prints global coordinate bounds: (x=-17.35, y=0, w=115.85, h=115.85);
trace('local coordinate bounds: ' + b.getBounds(b));//prints local coordinate bounds: (x=0, y=0, w=100, h=100)
HTH,
George
Chip's answer in code:
// convert degrees to radians
var r:Number = this.rotation * Math.PI/180;
// cos, c in the equation
var c:Number = Math.abs(Math.cos(r));
// sin, s in the equation
var s:Number = Math.abs(Math.sin(r));
// get the unrotated width
var w:Number = (this.width * c - this.height * s) / (Math.pow(c, 2) - Math.pow(s, 2));